Saturday, January 16, 2016

Geometry Problem 1180: Quadrilateral, 120 Degrees, Diagonals, Perpendicular

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1180.

Online Math: Geometry Problem 1180: Quadrilateral, 120 Degrees, Diagonals, Perpendicular.

Posted by Antonio Gutierrez at 7:52 AM
Labels: , , , ,

9 comments:

  1. Sumith PeirisJanuary 16, 2016 at 10:07 AM

    Let AB, DC meet at H so that E is the incentre of Tr. AHD and < BGE = < DGE = 30.

    Also easily < ACD = 60 + A/2 = < DBG ( A + 60 - A/2) making BECH concylic

    Hence BE = CE = 2FE since Tr . ECF is 30-60-90.

    Further Tr. s GED is congruent to Tr. CED case ASA making EG = EC

    ReplyDelete
    Replies
    1. Peter TranJanuary 16, 2016 at 11:46 AM

      I am not sure from "E is the incentre of Tr. AHD" , you get " < BGE = < DGE = 30.
      please explain .

      Peter

      Delete
    2. Sumith PeirisJanuary 16, 2016 at 3:31 PM

      Just a typo replace < BGE = < DGE = 30 with < BHE = < DHE = 30

      Delete
  2. Sumith PeirisJanuary 16, 2016 at 10:09 AM

    I could not complete the last line

    Hence EG = 2FE

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. Peter TranJanuary 16, 2016 at 11:13 AM

    http://s1.postimg.org/kgkafchkf/pro_1180.png

    BA meet CD at M
    Since ∠(A)+ ∠(D)= 120 =>∠(M)=60
    Since EA and ED bisect ∠(A) and ∠(D) => E is in center if trị. MAD and ME bisect ∠((M) and ∠( AED)=120
    MBEC is cyclic and chord BE=chord EC => trị BED is isosceles
    Triangle EFC is 30-60-90 triangle => EC= 2. EF
    Triangles ECD and EDG are congruence…. ( case ASA) => EG=EC=2. EF

    ReplyDelete
  4. c.t.e.oJanuary 17, 2016 at 3:08 AM

    To Sumith & Peter
    Why Tr CED congrc to EGD

    ReplyDelete
  5. Sumith PeirisJanuary 17, 2016 at 5:28 AM

    ASA. ED common. <GDE = <CDE and < GED = < CED = 60 which I'm sure u can figure out why.

    ReplyDelete
  6. c.t.e.oJanuary 17, 2016 at 7:18 AM

    OK to Sumith
    From congr in tr BGC, GF median and BM median (M on GC)
    =>EG = 2FE

    ReplyDelete
  7. Sumith PeirisJanuary 18, 2016 at 4:02 AM

    Interestingly Tr. BGC is equilateral from which the result EG = 2 FE is obvious.

    ReplyDelete

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